Two substances A $(t_{\frac{1}{2}} =$ 5 min) and B $(t_{\frac{1}{2}} =$ 15 min) are taken in such a way that initially $[A] = 4 [B]$ . The time after which both the concentrations will be equal is:

5 minutes

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15 minutes

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20 minutes

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concentrations can never be equal

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Solution

The correct option is B 15 minutes
C $_{t}$ = C $_{0}$ e $^{- K t}$

According to the question, $C_{A_{t}} = C_{B_{t}}$
$C_{A} e^{- K_{A} t} = C_{B} e^{- k_{B} t}$
$\frac{C_{A}}{C_{B}} = \frac{e^{- k_{A} t}}{e^{- K_{A} t}} \Rightarrow \frac{C_{A}}{C_{B}} = e (K_{A} - K_{B}) t$
$4 = e [\frac{l n 2}{5} - \frac{l n 2}{15}] \times t$
$l n 4 = [\frac{l n 2}{5} - \frac{l n 2}{15}]$ t
$l n (2)^{2} = [\frac{l n 2}{5} - \frac{l n 2}{15}]$ t
$2 l n 2 = [\frac{l n 2}{5} - \frac{l n 2}{15}]$ t
$2 = [\frac{1}{5} - \frac{1}{15}]$ t
$2 = \frac{2}{15} \times t$

$t = 15 m i n u t e s$

Hence, option B is correct.

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Similar questions

A (t_{\frac{1}{2}} = 5 min)

B (t_{\frac{1}{2}} = 15 min)

[A] = 4 [B]

A (t_{\frac{1}{2}} = 5 min)

B (t_{\frac{1}{2}} = 15 min)

[A] = 4 [B]

A

(t_{1 / 2} = 5 m i n s)

B

(t_{1 / 2} = 15 m i n s)

[A] = 4 [B]

A (_{1 / 2} = 5 m i n s)

t_{1 / 2} = 15 m i n s)

[A_{0}] = 4 [B_{0}]

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